abstract-algebrafield-theorypolynomials
Let $f=x^6-2$, find the degree of splitting field of $f$ over $\mathbb{Q}$.
I calculated the roots of $f$ are $\pm \sqrt[6]{2},\pm e^{i \pi/3}\sqrt[6]{2},\pm e^{2i \pi/3}\sqrt[6]{2}$. I suspect that the splitting field of $f$ is $\mathbb{Q}(e^{i \pi/3}\sqrt[6]{2})$ but I don't know how to prove this. Any idea?
Let $\zeta$ be a third root of unity and $K = \mathbf{Q} ( \sqrt[3]{5}), \zeta)$. $K$ contains the cyclotomic field of $3$rd roots of unity and is generated over it by $\sqrt[3]{5}$. Thus the degree is at most 6. $\mathbf{Q} (\sqrt[3]{5})$ and $\mathbf{Q}(\zeta)$ are both subfields of degrees 3 and 2 respectively. Since the degrees are relatively prime, the degree must be divisible by $3*2 = 6$ and so $[K: \mathbf{Q}] = 6$.
An even more direct route is to just use the fact that it is a composite field of relatively prime degrees.
Your final query regarding the Galois group is a bit confusing to me. Gal represents a Galois group, so it doesn't make sense to talk about its degree. I am guessing you mean to say order of the group then. In that case, this is always true. By definition, $K/F$ is Galois extension if $|\textrm{Aut} (K/F) | = [K:F]$ and we denote $\textrm{Aut} (K/F)$ as $\textrm{Gal}(K/F)$
Let $\alpha$ be a root of $x^4+1$. Then $\alpha^3,\,\alpha^5,\,\alpha^7$ are also (distinct!) roots of $x^4+1$. An easy way to see this is to consider 8th roots of unity, i.e. roots of $x^8-1 = (x^4+1)(x^4-1)$. If $\alpha$ is a primitive 8th root of unity, then every odd power $\alpha^{2k+1}$ is a root of $x^4+1$ (just draw a picture). Or you could just check it directly.
This means that $\mathbb Q[\alpha]$ is the splitting field of $x^4+1$. All you have to do now is to prove that $x^4+1$ is irreducible over $\mathbb Q$ to conclude that the degree of the splitting field is $4$.
EDIT: Perhaps a better way to show that $[\mathbb Q[\alpha]:\mathbb Q] = 4$ is to first notice that since $\alpha$ is a root of $x^4+1$ that $[\mathbb Q[\alpha]:\mathbb Q] \leq 4$. Now notice that $\alpha + \alpha^7 = \sqrt 2$, so $\mathbb Q[\sqrt 2]\subseteq \mathbb Q[\alpha]$. But, $\mathbb Q[\sqrt 2]\subseteq\mathbb R$, while $\alpha$ is complex, so $\mathbb Q[\sqrt 2]\neq \mathbb Q[\alpha]$, so it must be that $[\mathbb Q[\alpha]:\mathbb Q] = 4$.
In the bellow graph are shown 8th roots of unity. Red are roots of $x^4+1$ and blue are roots of $x^4-1$.
Best Answer
The splitting field of $f$ is the field $\mathbf Q(\zeta,\sqrt[6]2)$, where $\zeta$ is a primitive $6$th root of unity. The minimal polynomial of $\zeta$ is the cyclotomic polynomial $\Phi_6(x)=x^2-x+1$, so that $$[\mathbf Q(\zeta):\mathbf Q]=2, \enspace[\mathbf Q(\sqrt[6]2):\mathbf Q]=6.$$ Furthermore, these extensions are linearly independent and thus $$[\mathbf Q(\zeta\sqrt[6]2):\mathbf Q]=[\mathbf Q(\zeta):\mathbf Q][\mathbf Q(\sqrt[6]2:\mathbf Q]=12.$$